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Calculate $\sum\limits_{i=2}^\infty\sum\limits_{j=2}^\infty \frac{1}{j^i}$. I am trying to figure out how to calculate this. I know it must be $\lim_{k\rightarrow\infty}\sum\limits_{i=2}^k\sum\limits_{j=2}^k\frac{1}{j^i}$ but I am not sure how to do this?

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  • $\begingroup$ Hint: sum over $i$ first. $\endgroup$ Commented Mar 5, 2016 at 17:39
  • $\begingroup$ So $\frac{1}{j^2}+\frac{1}{j^3}+\frac{1}{j^4}+\frac{1}{j^5}$... $\endgroup$
    – user320195
    Commented Mar 5, 2016 at 17:41

3 Answers 3

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Hint (as first proposed by achille hui in the comments above): $$\sum_{i=2}^\infty j^{-i}=\frac{1}{j(j-1)} \; , \;\;j>1$$

EDIT:

The above is a geometric series $\sum_{i=0}^\infty r^i=\frac{1}{1-r}$ with $r=\frac{1}{j}$ where you have to subtract the first two terms (since $i$ runs from $2$ and not $0$).

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    $\begingroup$ You should also argue that exchanging the order of summation doesn't change the limit - this is not generally true for doubly infinite sequences, but is true when the series is absolutely convergent (as it is here), by Fubini's theorem. $\endgroup$
    – user93238
    Commented Mar 5, 2016 at 17:51
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    $\begingroup$ @Mark For this case, it is easier to use Tonelli (as all terms involved are obviously non-negative). $\endgroup$ Commented Mar 5, 2016 at 17:57
  • $\begingroup$ @Mark Since the summands are nonnegative, its allowed even if the sums are infinite. $\endgroup$
    – Pedro
    Commented Mar 5, 2016 at 18:02
  • $\begingroup$ Thank you for all the input! Why is it that $\sum\limits_{i=2}^\infty j^{-1}=\frac{1}{j(j-1)}$? $\endgroup$
    – user320195
    Commented Mar 5, 2016 at 18:12
  • $\begingroup$ @user320195 it is a geometric series. $\endgroup$ Commented Mar 5, 2016 at 18:19
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$$ \begin{align} \sum_{i=2}^\infty\sum_{j=2}^\infty\frac1{j^i} &=\sum_{j=2}^\infty\sum_{i=2}^\infty\frac1{j^i}\tag{1}\\ &=\sum_{j=2}^\infty\frac{\frac1{j^2}}{1-\frac1j}\tag{2}\\ &=\sum_{j=2}^\infty\frac1{j(j-1)}\tag{3}\\ &=\sum_{j=2}^\infty\left(\frac1{j-1}-\frac1j\right)\tag{4}\\ &=\lim_{n\to\infty}\sum_{j=2}^n\left(\frac1{j-1}-\frac1j\right)\tag{5}\\ &=\lim_{n\to\infty}\left(1-\frac1n\right)\tag{6}\\[12pt] &=1\tag{7} \end{align} $$ Explanation:
$(1)$: since all terms are positive, apply Tonelli's Theorem
$(2)$: use the Formula for the Sum of a Geometric Series
$(3)$: simplify the summand
$(4)$: apply Partial Fractions
$(5)$: apply the definition of an Infinite Series
$(6)$: compute the Telescoping Sum
$(7)$: evaluate the limit

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Notice:

  • $$\sum_{n=a}^{\infty}\frac{b}{n^c}=b\zeta(c,a)\space\text{ when }b=0\vee\Re(c)>1$$
  • $$\sum_{n=a}^{\infty}\frac{b}{c^n}=\frac{bc^{1-a}}{c-1}\space\text{ when }|c|>1$$

So, to solve your question:

$$\sum_{n=2}^{\infty}\sum_{m=2}^{\infty}\frac{1}{m^n}=\left[\sum_{n=2}^{\infty}\left[\sum_{m=2}^{\infty}\frac{1}{m^n}\right]\right]=\left[\sum_{n=2}^{\infty}\left[\zeta(n)-1\right]\right]=1$$

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